Polynomials and the Generalized Mean Value Theorem I want to prove the following proposition:
Proposition. There exist real polynomials $f$ and $g$ such that $g(0)\neq g(1)$ and $$ \frac{f(1)-f(0)}{g(1)-g(0)}\neq \frac{f^{'}(\xi)}{ g^{'}(\xi)}$$ for all $\xi\in (0,1)$.
It seems directly related to the generalized mean value theorem (MVT) but am not sure how to prove it. The negation of the statement is:
For all real polynomials $f$ and $g$ such that $g(0)\neq g(1)$ there exist $\xi\in (0,1)$ such that $$ \frac{f(1)-f(0)}{g(1)-g(0)}= \frac{f^{'}(\xi)}{ g^{'}(\xi)}$$
which is true if $g^{'}(\xi)\neq 0$ by the generalized MVT. Any help is greatly appreciated.
 A: What you can prove is that (in particular) for every real polynomial functions  $f$,$g$ 
such that $g(0) \neq g(1)$ you can find a $c \in (0,1)$ such that
$$
\left[ {f(1) - f(0)} \right]g'(c) = \left[ {g(1) - g(0)} \right]f'(c)
$$
thus either $g'(c)=0$ so that $f'(c)=0$ also, or 
$$
\frac{{f(1) - f(0)}}
{{g(1) - g(0)}} = \frac{{f'(c)}}
{{g'(c)}}
$$
Hence you can build an example of $f$ and $g$ where there exists a $c \in (0,1)$ such that $f'(c)=g'(c)=0$ and such that for every other $\xi \in (0,1)$ with $\xi \neq c$ it is 
$$
\frac{{f(1) - f(0)}}
{{g(1) - g(0)}} \ne \frac{{f'(\xi )}}
{{g'(\xi )}}
$$
For instance you can choose
$$
f(x) = \left( {x - \frac{1}
{2}} \right)^3  + \left( {x - \frac{1}
{2}} \right)^2 
$$
$$
g(x) = \left( {x - \frac{1}
{2}} \right)^3  - \left( {x - \frac{1}
{2}} \right)^2 
$$
You have that
$$
\frac{{f(1) - f(0)}}
{{g(1) - g(0)}} = 1
$$
$f'(1/2)=g'(1/2)=0$ and for every $\xi \neq 1/2$ it is
$$
\frac{{f'(\xi )}}
{{g'(\xi )}} = \frac{{3\left( {\xi  - \frac{1}
{2}} \right) + 2}}
{{3\left( {\xi  - \frac{1}
{2}} \right) - 2}}
$$
which, of course , can never be $1$.
A: Let $f(x)=(x-\frac13)^3, g(x)=(x-\frac13)^2$. If
$$ \frac{f(1)-f(0)}{g(1)-g(0)}= \frac{f^{'}(\xi)}{ g^{'}(\xi)}$$
one has
$$ \frac{(\frac23)^3-(-\frac13)^3}{(\frac{2}{3})^2-(\frac13)^2}=\frac{3(\xi-\frac13)^2}{2(\xi-\frac13)}$$
which has solution $\xi=1$. But $\xi=1$ is not in $(0,1)$.
